Find the roots of the quadratic equation $3x^{2}-2\sqrt{6}x+2=0$.

(N/A) The given quadratic equation is $3x^{2}-2\sqrt{6}x+2=0$.
We can factorize the middle term $-2\sqrt{6}x$ as $-\sqrt{6}x - \sqrt{6}x$.
So,$3x^{2}-\sqrt{6}x-\sqrt{6}x+2=0$.
Taking common terms from the first two and last two terms:
$\sqrt{3}x(\sqrt{3}x-\sqrt{2})-\sqrt{2}(\sqrt{3}x-\sqrt{2})=0$.
This simplifies to $(\sqrt{3}x-\sqrt{2})(\sqrt{3}x-\sqrt{2})=0$.
Setting each factor to zero,we get $\sqrt{3}x-\sqrt{2}=0$,which implies $x = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}}$.
Since both factors are identical,the roots are equal.
Therefore,the roots of the equation are $\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}}$.